1 introduction to communications professor r. c. t. lee dept. of information management dept. of...

1

Introduction to Communications

Professor R. C. T. Lee

Dept. of Information Management

Dept. of Computer Science

Department of Communications

Department of Live Science

National Chi Nan University

2

• Why do we computer scientists need to know communication technologies?

3

• In these days, it is hard to imagine any computer which stands entirely alone. A computer is often connected to others, through communication technologies.

4

• Besides, many systems which were traditionally considered as communication systems are actually quite similar to computers.

• Examples: TV, mobile phone.

5

• But, unfortunately, almost all computer scientist students do not understand some very basic communication technologies.

• Question 1: How is a bit represented when it is transmitted in a wireless environment?

6

• We often thought that a bit is represented by a pulse.

• (1,1,0,0,1)

1 1 10 0

7

• But we cannot send a digital signal in the wireless environment.

• Question 2: We often mix bits together.

• An example: In the ADSL system, 256 bits are bundled together.

8

• If the bits are represented by pulses, how can they be mixed and later be distinguished.

• Question 3: We often mix bits by different users together.

• If the bits are represented by pulses, how can they be distinguished when they are received?

9

• A very fundamental problem in communications is to study how digital information is represented, sent and later separated.

• Fundamental concept: There is no digital signal. Every digital bit is represented by an analog signal. (A quote by Professor R. C. T. Lee, a famous professor. He learned this from one of his students.).

10

• Consider the simplest case: There is only one user.

• We represent bit 1 by and represent bit 0 by .

• We further require that and are orthgonal.

•

)(1 tf)(2 tf

)(1 tf )(2 tf

11

• That is, the inner product

• and for every i. .

0)()()(),(0 2121 dttftftftfT

1)(),( tftf ii

12

For the receiver, it either receives or .

We perform two inner products.

Let the sent signal, which is received denoted by .

We calculate and

If , we conclude that we sent , which is 1(0).

)(1 tf )(2 tf

)(ts

)(),( 11 tftsx .)(),( 22 tftsx

)1(1 21 xx

))()(( 21 tftf

13

There are millions of possible functions for and .

We may let and

We may also let and

We may of course let and .

)(1 tf )(2 tf

)2cos()(1 tftf c ).2sin()(2 tftf c

)2cos()(1 tmftf c

).2cos()(2 tnftf c

)2sin()(1 tmftf c)2sin()(2 tnftf c

14

It can be easily seen that

We can also prove, for instance, that

.0)2sin()2cos()2sin(),2cos(0

dttftftftf c

T

ccc

0

)2cos()2cos(

)2cos(),2cos(

0

dttnftmf

tnftmf

cc

T

cc

15

What we usually do is to find If we conclude that the sent signal is

Or, if we conclude that the sent bit is 1(0).

.21 xxy

),0(0 yy)).()(( 21 tftf

),0(0 yy

)(),( 11 tftsx .)(),( 22 tftsx and

11 x iff )()( 1 tfts

12 x iff )()( 1 tfts

16

We should now always remember that everybit, 1 or 0, is represented by a cosine ora sine function.

In the above, we assumed that every bit is sent alone.

Can we send two bits together?

Yes, we can.

17

Let us assume that we are going to send twobits: Bit 1 and Bit 2.

Bit 1 can be 1 or 0 and Bit 2 can also be 1 or0.

We like to mix Bit 1 and Bit 2 together and send the mixed signal out.

The important thing is that the receiver mustbe able to correctly determine what Bit 1 and Bit 2 are .

18

Let Bit 1(Bit 2) be represented by .

The value of is determined by the value of Bit 1(Bit 2) .

We may let be 1(-1) if Bit i is 1(0) for

Thus the sent signal is

The job of the receiver is to determine the values of for

)).()(( 2211 tfatfa

)( 21 aa

ia .2,1i

).()( 2211 tfatfa

ia .2,1i

19

Let denote the sent signal. Then

To determine , we perform an inner Product between and

Let

)(ts

).()()( 2211 tfatfats

,1a

)(ts ).(1 tf

.)(),( 11 tftsx

20

.

)(),()(),(

)(),()(

)(),()(

1

122111

12211

11

a

tftfatftfa

tftfatfa

tftstx

Similarly, we have

.

)(),()(

)(),()(

2

22211

22

a

tftfatfa

tftstx

21

If we conclude that User 1 sends 1(0).

Similarly, if we conclude that User 2sends 1(0).

),1(1 11 aa

),1(1 22 aa

22

We can now see how two bits can be mixedand sent without any trouble.

Essentially, we must understand that thedigital signals are represented by analog signals which are orthogonal to each other.

The receiver uses the property of orthogonalityto separate the signals.

23

If we can mix 2 bits together, we can of Course mix 256 bits together.

In fact, the ADSL system uses this kind of scheme.

This is why the ADSL system is a very fast system.

24

In the ADSL system, each bit i is representedby

Thus, each signal is orthogonal to others due to the distinct frequencies.

This is why the system is called OrthogonalFrequency Division Multiplexing (OFDM) system.

.256,,2,1),2cos( itfi

25

The coding of digital data by analogsignals is often called digital modulation.

The decoding of analog signals back todigital signals is called demodulation.

26

Let us go one step further by mixing bitsfrom different people.

Our trick is the following: A bit of 1 or 0 forUser 1 is represented differently from aBit of 1 or 0 for User 2.

Let us consider a simple case: two users.

27

Let the bit of 1 for User 1 be coded as and the bit of 0 for User 1 be coded as

Let the bit of 1 for User 2 be coded as and the bit of 0 for User 2 becoded as

We may say that User i sends

)1,1(1 s

).1,1()1,1(1 s

)1,1(2 s).1,1(2 s

.iisa

28

Since and are vectors, we use dot-product as the inner product.

We therefore conclude that and areorthogonal to each other.

1s 2s

0

)1(1

))1(1()11(

)1,1()1,1(

, 21

ss

1s 2s

29

We may mix the bits of User 1 and User 2and the mixed signal is therefore where or .

The job of the receiver is to determine

Again, this can be done by using the orthogonality of and

,2211 sasa 1ia 1

.ia

1s .2s

30

To find we calculate

Thus

can be found similarly.

,1a

.2

,,

,

,

1

122111

12211

11

a

ssassa

ssasa

ssx

.21

1

xa

2a

31

• Thus we can see that we can even mix

• bits of two different users.

• The main trick is that we represent the

• bits of different users by vectors and make

• sure that they are orthogonal to each other.

• This can be easily extended to more than 2 users.

32

Let us now ask another interesting question.

Our cables are often used to transmit digitaldata.

We like our channel to transmit a large amount of bits per second.

We are talking about high bit rate systems.

Why do we often call these channelsbroadband system.

33

By

By a broadband system, we mean it can transmit signals with a large range of frequencies .

Why is a high bit system also a broadbandsystem?

This can be understood only through theknowledge of Fourier transform.

34

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-4

-3

-2

-1

0

1

2

3

4

Time

t

f(t)

A signal

35 The Discrete Fourier Transform Spectrum of the

Signal in the Previous Slide

0 10 20 30 40 50 600

0.5

1

1.5

2

2.5

3

3.5

4

Frequency

f

36A Music Signal Lasting 1 Second

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

Time

t

37

0 2000 4000 6000 8000 10000 12000 14000 160000

50

100

150

Frequency

f

A Discrete Fourier Transform Spectrum of the Music Signal in the Previous Slide.

38

Suppose that a bit is a rather wide one, asshown below:

Then its Fourier transform spectrum is as follows:

39

Suppose the system is a high bit rate system,the bit length is therefore very very short:

Its Fourier transform is as follows:

40

Conclusion: In a low bit rate system, the bitlength is very long and it actually contains anarrow band of frequencies.

This can be understood by imagining thebit length to be so long that the signalbecomes DC.

In this case, the only frequency it containsis .0f

41

On the contrary, in a high bit rate system, thebit length is very very short, it contains a widerange of frequencies.

The system is consequently called abroadband system because for a wide range of frequencies, it must respond equally well.

42

• By using Fourier transform, we can see that the frequency components in our human voice are roughly contained in 3k Hertz.

43

• For a signal with frequency f, its wavelength can be found as follows:

• where f is the velocity of light?

f

v

44

• If , .3103f km100m10103

103 53

8

45

• It can also be proved that the

length of an antenna is around .

• For human voice, this means that the wavelength is 50km.

• No antenna can be that long.

2

46

• What can we do?

• Answer: By amplitude modulation.

47

• Let be a signal.

•

The amplitude modulation is defined as follows:

• where fc is the carrier frequency?

)(tx

)2cos()( tftx c

48

• What is the Fourier transform of

? )2cos()( tftx c

49

50

Fourier transform tells us that every signal containsa bunch of cosine functions.

Let us consider .)2cos( ft

))(2cos())(2(cos(2

1

)2cos()2cos(

tfftff

tfft

cc

c

51

Thus every frequency f is lifted to

.ffandff cc

))(2cos())(2(cos(2

1

)2cos()2cos(

tfftff

tfft

cc

c

52

• The effect of amplitude modulation is to lift the baseband frequency to the carrier frequency level, a much higher one.

• Once the frequency becomes higher, its corresponding wavelength becomes smaller.

• An antenna is now possible.

53

• After we receive , how can we take out of it?

• Answer: Multiply by .

)2cos()()( tftxts c)(tx

)(ts )2cos( tfc

54

• Thus is recovered.

)2(cos)()2cos()( 2 tftxtfts cc

)))2(2cos()()((2

1tftxtx c

)(tx

55

56

• Thank You for Your Patience.

• Good-night and Have a Good Sleep.

• You will be waken up by Your Wicked Advisers Tomorrow Morning.